Trying to understand the very basics of modules of differentials (Kähler differentials) So I am just learning about modules of differentials at the moment, and am not seeing some of the basic properties. I wanted to see if someone could run through the calculations of the most basic explicit example so I can see what's going on behind the scenes. Say I have the polynomial ring $A = \mathbb{C}[x, y]$ in two variables as a $\mathbb{C}$-algebra. I want to construct the module of differentials $\Omega_{A / \mathbb{C}}$ and then show that it is isomorphic to a free module generated by $dx$ and $dy$. I define $\Omega_{A / \mathbb{C}}$ as the free module generated by the set $\{ df: f \in \mathbb{C}[x, y] \}$ (call this free module $M$) subject to the following relations
\begin{align}
d(f+g) &= df + dg \\
d(fg) &= fdg + gdf \\
d(\alpha) & = 0 \quad \alpha \in \mathbb{C}
\end{align}
So far this is standard and intuitive. However the (seemingly) most basic first claim is that this is isomorphic to the free module on $dx$ and $dy$. Call this free module $N$. I was wondering if someone could spell this out in explicit detail. In particular, I know I want a surjective map
$$
\phi: M \longrightarrow N,
$$
so that the kernel is precisely the ideal generated by the relations above. What is this surjective map exactly, and why is the kernel given by those relations? I feel like this should be extremely simple, but I am obviously not seeing something.
 A: Well, each $f$ is the sum of monomials $ax^r y^s$ and
the relations in $\Omega=\Omega_{A/\Bbb C}$ means that
$$d(ax^ry^s)=arx^{r-1}y^s\,dx+asx^ry^{s-1}\,dy$$
exactly as you expect. 
So we get a $\Bbb C$-linear map from $M$
to $N$ with
$$d\left(\sum_{r,s}a_{r,s}x^ry^s\right)\mapsto
\sum_{r,s}a_{r,s}\left(rx^{r-1}y^s\,dx+sx^r y^{s-1}\,dy\right).$$
This $d$ is a $\Bbb C$-linear derivation: $d(fg)=f\,dg+g\,df$.
Obviously onto, but what about verifying the kernel is as you claim.
Well, one does not do that directly, but rather one shows that
$N$ has the right universal property. The correct universal property is
that $d:A\to N$ given by $f\mapsto df$
with the above definition of $d$ is a universal derivation. This
means that if $\delta:A\to P$ is a $\Bbb C$-linear map to an
$A$-module $P$ with $\delta(fg)=f\delta(g)+g\delta(f)$ then there
is a unique $A$-module map $\psi:N\to P$ with $\delta=\psi\circ d$.
To define this $\psi$, it suffices to determine where $dx$ and $dy$
go (as $N$ is free on these), and of course they must go to $\delta(x)$
and $\delta(y)$. It's routine to verify that this $\psi$ satisfies
$\psi(df)=\delta(f)$.
